An object is lunched at a speed of 25 m/s from the top of a tall tower with cert

Question

An object is lunched at a speed of 25 m/s from the top of a tall tower with certain angle theta, The height y of the object as a function of the time t elapsed from the lunch is y(t) = -4.9^t2 + 18.5t + 65, where h is in meters and t in seconds. Determines the height of the tower; the lunch angle; the horizontal distance travel by the object before it hits the ground. Suppose you decide to drop a melon from rest from the first observation platform of the Eiffel Tower. The initial height h from which the melon is released is 68.3 m above the head of your French friend Pierre, who is standing on the ground right below you. At the same instant you release the melon, Pierre shoots an arrow straight up with an initial velocity of 27.0 m/s. (Of course, Pierre makes sure the area around him is cleared and gets out of the way quickly after he shoots his arrow.) How long after you drop the melon will the arrow hit it? At what height above Pierre's head does this collision occur?

Explanation / Answer

H = Hi + Vi t + a t² / 2
where
H = height (at time t)
Hi = initial height (at time 0)
Vi = initial vertical velocity (at time 0)
t = time
a = acceleration by gravity = -9.8 m/s²

Now, the formula you were given, is exactly the same, but with some values already filled in.
I'll rearrange it a bit so that you can clearly see it.

y(t) = 4.9t² + 18.5*t + 65
y(t) = 65 + 18.5*t 4.9t²

H = Hi + Vi t + a t² / 2

so

y(t) = H = height (at time t)
65 = Hi = initial height (at time 0)
18.5 = Vi = initial vertical velocity (at time 0)
t = time
-4.9 = a/2 = (acceleration by gravity)/2 => a = -9.8 m/s²

"Determine the height H of the tower"
That would be Hi, so
height of the tower is 65 m < - - - - - - - - - - - - - - - - - - - Answer height tower

"Determine the launch angle"
sin() = Vy / Vi
where
= launch angle = ?
Vy = initial vertical velocity = 18.5 m/s
Vi = initial actual velocity = 25.0 m/s
so
sin() = 18.5/25
= arcsin(0.74)
= 47.73° < - - - - - - - - - - - - - - - - - - - Answer launch angle

"Determine the horizontal distance traveled by the object before it hits the ground."
First find the time the object needs to hit the ground.
H = Hi + Vi t + a t² / 2
where
H = height = 0 m (you want the time when height is 0m)
Hi = initial height = 65 m (height of the tower)
Vi = initial vertical velocity = 18.5 m/s
t = time = ?
a = acceleration by gravity = -9.8 m/s²
so
0 = 65 + 18.5*t + (-9.8)t²/2
0 = 65 + 18.5*t - 4.9t²
(this is your original equation where you say that y(t) must be 0m)

A negative time makes no sense here, so there's only one valid solution
t = 6.0 s

Now find the horizontal component of the inital velocity.
Vh = Vi cos()
where
Vh = horizontal component = ?
Vi = initial velocity = 25 m/s
= launch angle = 47.73°
so
Vh = 25*cos(47.73)
Vh = 16.82 m/s

And finaly, find the horizontal range
R = Vh t
where
R = horizontal range = ?
Vh = horizontal velocity = 16.82 m/s
t = 6.0 s
so
R = 16.82*6.0
R = 100.92 m < - - - - - - - - - - - - - - - - - - - Answer horizontal range

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